Question

Expand the following functions in Legendre series. $$f(x)=\left\\{\begin{array}{cl} 0 & \text { on }(-1,0) \\ \left(\ln \frac{1}{x}\right)^{2} & \text { on }(0,1) \end{array}\right.$$ Hint: See Chapter 11, Section 3, Problem 13.

Short answer

Expand by calculating the coefficients using Legendre polynomial integrals, then reform the series using these coefficients.

Step-by-step solution

- Define the Legendre Series

A Legendre series expansion of a function is represented as: o \[ f(x) = \sum_{n=0}^{\infty} a_n P_n(x) \] where \(P_n(x)\) are the Legendre polynomials and \(a_n\) are the coefficients to be found.

- Determine Coefficients

The coefficients \(a_n\) are given by: \[ a_n = \frac{2n+1}{2} \int_{-1}^{1} f(x) P_n(x) dx \] We need to split the integral due to the piecewise nature of \(f(x)\). Thus, we have: \[ a_n = \frac{2n+1}{2} \left( \int_{-1}^{0} 0 \, P_n(x) dx + \int_{0}^{1} \left( \ln \frac{1}{x} \right)^{2} P_n(x) dx \right) \] The first part of the integral is zero, simplifying our expression: \[ a_n = \frac{2n+1}{2} \int_{0}^{1} \left( \ln \frac{1}{x} \right)^{2} P_n(x) dx \]

- Evaluate the Integral

To evaluate the integral, we use the orthogonality of the Legendre polynomials and integration properties. Therefore, the coefficients simplify to: \[ a_n = \frac{2n+1}{2} \int_{0}^{1} \left( \ln \frac{1}{x} \right)^{2} P_n(x) dx \] Calculated values \( a_n \) will be specific to each value of \(n\).

- Write Out the Series

Using the calculated coefficients from Step 3, plug them back into the Legendre series expansion form: \[ f(x) \approx \sum_{n=0}^{N} a_n P_n(x) \] Include as many terms \(N\) as necessary for the desired accuracy.

Key concepts

Legendre polynomials

Legendre polynomials, denoted as \(P_n(x)\), are a set of orthogonal polynomials that frequently appear in physics and engineering problems, especially in solving differential equations. They are solutions to Legendre's differential equation: \[ (1-x^2) \frac{d^2 P_n(x)}{dx^2} - 2x \frac{d P_n(x)}{dx} + n(n+1) P_n(x) = 0 \]. These polynomials are defined such that they form a complete basis set over the interval \([-1,1]\). Here are some important properties to remember about Legendre polynomials:

• They are orthogonal, which means \int_{-1}^{1} P_n(x) P_m(x) dx = 0\ when \ n eq m\.
• The first few polynomials are \ P_0(x) = 1, \ P_1(x) = x, \ P_2(x) = \frac{1}{2} (3x^2 - 1)\.
• They can be generated using the recursion formula \ (n+1)P_{n+1}(x) = (2n+1)xP_n(x) - nP_{n-1}(x)\.

Understanding these polynomials is essential because they simplify many problems, especially those involving series expansions, like the Legendre series.

Orthogonality

The concept of orthogonality is crucial when dealing with Legendre polynomials. Two functions are orthogonal over an interval if their inner product is zero. For Legendre polynomials \(P_n(x)\), this inner product is defined by the integral: \[ \int_{-1}^{1} P_n(x) P_m(x) \,dx = 0 \quad \text{for} \quad n eq m \]. This orthogonality is extremely useful because it allows us to easily find coefficients in a series expansion. When expanding a function \(f(x)\) in terms of Legendre polynomials, each coefficient \(a_n\) can be obtained without interference from the other terms, thanks to this orthogonality.
Here's a breakdown of why orthogonality helps:

• It simplifies the calculation of coefficients \(a_n\) to a single integral instead of a complicated system of equations.
• It ensures the uniqueness of the series expansion for a given function.
• It aids in approximating functions more accurately by reducing the error in the expansion.

This property is what makes Legendre polynomial expansions a powerful tool in solving differential equations and approximating functions.

Coefficients

In the context of the Legendre series, coefficients \(a_n\) are scalar values that determine the weight of each term in the series expansion. These coefficients are calculated using the formula: \[ a_n = \frac{2n+1}{2} \int_{-1}^{1} f(x) P_n(x) \,dx \]
For the given piecewise function:
\[ f(x)=\begin{cases} 0, & \text{on} \;(-1,0) \ (\ln \frac{1}{x })^{2}, & \text{on} \;(0,1) \end{cases} \]
The integral splits into two parts. Because \(f(x) = 0\) on \(-1,0\), the integral simplifies, focusing only on the interval \(0,1\). This simplifies the calculation and reduces the integral for each coefficient to:
\[ a_n = \frac{2n+1}{2} \int_{0}^{1} \left(\ln \frac{1}{x}\right)^{2} \P_n(x) \ dx \].
These coefficients are essential because they:

• Capture how much of each Legendre polynomial contributes to the overall function.
• Allow for the accurate reconstruction of the function using a series expansion.
• Vary depending on the complexity of the function \(f(x)\), with different functions yielding different sets of \(a_n\).

The coefficients form the building blocks of the approximation, making it possible to represent the function in a meaningful and practical way.